A) \[\frac{1}{8}\]
B) \[\frac{3}{16}\]
C) \[\frac{\sqrt{3}}{16}\]
D) \[\frac{1}{16}\]
Correct Answer: D
Solution :
\[\sin 10{}^\circ \cdot \sin 30{}^\circ \cdot \sin 50{}^\circ \cdot \sin 70{}^\circ \] \[=\frac{1}{2}\cdot \sin 10{}^\circ \cdot \frac{1}{2}(2\sin 70{}^\circ \cdot \sin 50{}^\circ )\] \[=\frac{1}{2}\cdot \sin 10{}^\circ \cdot \frac{1}{2}\{\cos (70{}^\circ -50{}^\circ )-\cos (70{}^\circ +50{}^\circ )\}\] \[=\frac{1}{2}\cdot \sin 10{}^\circ \cdot \frac{1}{2}\{\cos 20{}^\circ -\cos 120{}^\circ \}\] \[=\frac{1}{2}\sin 10{}^\circ \cdot \frac{1}{2}\left( \cos 10{}^\circ +\frac{1}{2} \right)\] \[=\frac{1}{2}\sin 10{}^\circ \cdot \cos 20{}^\circ +\frac{1}{8}\sin 10{}^\circ \] \[=\frac{1}{4}\cdot \frac{1}{2}(\sin 30{}^\circ -\sin 10{}^\circ )+\frac{1}{8}\cdot \sin 10{}^\circ \] \[=\frac{1}{8}\cdot \sin 30{}^\circ -\frac{1}{8}\sin 10{}^\circ +1/8\,\sin 10{}^\circ \] \[=\frac{1}{8}\cdot \frac{1}{2}-0=\frac{1}{16}\]
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